Question

pythagorean theorem
determine the missing length in each right triangle using the pythagorean theorem. round
the answer to the nearest tenth.
1)
2)
3)
su =
kl =
fg =
4)
5)
6)
rt =
ab =
nl =
7)
8)
9)

Explanation:

Step1: Recall the Pythagorean theorem

The Pythagorean theorem is \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse of a right - triangle and \(a\) and \(b\) are the other two sides. If we need to find one of the non - hypotenuse sides, we can rewrite the formula as \(a=\sqrt{c^{2}-b^{2}}\).

Step2: Solve for \(SU\) in the first triangle

In triangle \(STU\), \(ST = 7\) yd, \(TU=14\) yd. Using the Pythagorean theorem \(SU=\sqrt{ST^{2}+TU^{2}}=\sqrt{7^{2}+14^{2}}=\sqrt{49 + 196}=\sqrt{245}\approx15.7\) yd.

Step3: Solve for \(KL\) in the second triangle

In triangle \(JKL\), \(JK = 5\) in, \(JL = 10\) in. Using the Pythagorean theorem \(KL=\sqrt{JL^{2}-JK^{2}}=\sqrt{10^{2}-5^{2}}=\sqrt{100 - 25}=\sqrt{75}\approx8.7\) in.

Step4: Solve for \(FG\) in the third triangle

In triangle \(EFG\), \(EF = 4\) ft, \(EG = 3\) ft. Using the Pythagorean theorem \(FG=\sqrt{EF^{2}+EG^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{16 + 9}=\sqrt{25}=5\) ft.

Step5: Solve for \(RT\) in the fourth triangle

In triangle \(RTS\), \(RS = 6\) in, \(TS = 2\) in. Using the Pythagorean theorem \(RT=\sqrt{RS^{2}-TS^{2}}=\sqrt{6^{2}-2^{2}}=\sqrt{36 - 4}=\sqrt{32}\approx5.7\) in.

Step6: Solve for \(AB\) in the fifth triangle

In triangle \(ABC\), \(AC = 4\) ft, \(BC = 9\) ft. Using the Pythagorean theorem \(AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{4^{2}+9^{2}}=\sqrt{16 + 81}=\sqrt{97}\approx9.8\) ft.

Step7: Solve for \(NL\) in the sixth triangle

In triangle \(MNL\), \(MN = 15\) yd, \(ML = 25\) yd. Using the Pythagorean theorem \(NL=\sqrt{ML^{2}-MN^{2}}=\sqrt{25^{2}-15^{2}}=\sqrt{625 - 225}=\sqrt{400}=20\) yd.

Step8: Solve for \(YZ\) in the seventh triangle

In triangle \(XYZ\), \(XY = 11\) ft, \(XZ = 13\) ft. Using the Pythagorean theorem \(YZ=\sqrt{XZ^{2}-XY^{2}}=\sqrt{13^{2}-11^{2}}=\sqrt{169 - 121}=\sqrt{48}\approx6.9\) ft.

Step9: Solve for \(PR\) in the eighth triangle

In triangle \(PQR\), \(PQ = 8\) yd, \(QR = 3\) yd. Using the Pythagorean theorem \(PR=\sqrt{PQ^{2}-QR^{2}}=\sqrt{8^{2}-3^{2}}=\sqrt{64 - 9}=\sqrt{55}\approx7.4\) yd.

Answer:

1. \(SU\approx15.7\) yd
2. \(KL\approx8.7\) in
3. \(FG = 5\) ft
4. \(RT\approx5.7\) in
5. \(AB\approx9.8\) ft
6. \(NL = 20\) yd
7. \(YZ\approx6.9\) ft
8. \(PR\approx7.4\) yd